Calculating Power Dissipated by a Battery

[bensthebest]

Hey I am very new here (litrally signed up 10 mins ago after trying to search the net for this problem)

heres what it looks like in my c/w

A Battery with an open voltage of 12v and an internal resistance of 0.25\Omega is connected to the circuit terminals A and B shown below (A is the top terminal and B is the bottom i forgot to add it when i copied the picture with paint!)

so far I've worked out the total resistance of the open circuit but the next question is asking

c) the terminal voltage of the battery when it is connected to the circuit

i attempted to use ohms law but that would of been pointless. then i looked at the question again and it tells me it has 12v any way but i doubt they would have a question giving me the answer like that and then offer 2 marks for it.

The next question I am stuck on is

f) The total power dissipated by the battery

again i really have no idea how to work that out.

 

[Chi Meson]

The battery itself adds 0.25 ohms to the total resistance you have already found.

The 12V from the battery is the "emf," but the terminal voltage is the voltage drop outside the battery. Some of that 12 V is "used up" inside the battery by the internal resistance.

So find the total current using the total resistance (including the internal resistance), and then using that current find how much voltage is dropped across the external circuit (using just the resistance you found in the first part).

And look at the power equation and consider the current through the battery, and the resistance of the battery.

 

[perjogab]

Power= Current * Voltage

 

[bensthebest]

The battery itself adds 0.25 ohms to the total resistance you have already found.

The 12V from the battery is the "emf," but the terminal voltage is the voltage drop outside the battery. Some of that 12 V is "used up" inside the battery by the internal resistance.

So find the total current using the total resistance (including the internal resistance), and then using that current find how much voltage is dropped across the external circuit (using just the resistance you found in the first part).

And look at the power equation and consider the current through the battery, and the resistance of the battery.

but how can i work out the current as i only have resistance and the voltage is what i need to find?

im using R=V / I

so the total resistance including the battery is 5.5 ohms

so to find the current i use the 12v is gives me?

 

[perjogab]

Total resistance = 5.5 ohms. I=V/R
Current=12/5.5
Power = Current * Voltage
Power = 2.14*12
Power= 25.68J (approximation)

 

[bensthebest]

thank you for the answer but i really want to know the hows and whys.

 

[perjogab]

uhh, resistors in a parallel share the same voltage but have different current, while resistors in a serious have different voltage but the same current.
Formula for resistors in a parallel is Rtotal= 1/R1+1/R3+1/R3...+1/Rn
For resistors in a parallel you simply add them
FOR VOLTAGE IN A CIRCUIT
Vtotal=Itotal*Rtotal
Therefore, Itotal=Vtotal/Rtotal
You can transpose to find w/e else you may need from that.

 

[perjogab]

So like I said in the post above, in a series the voltage in the circuits are different. So to find the voltage in the 0.25 ohm resistor you would say once again:
V=IR
But in this case R is 0.25 ohms, I would be the current of the circuit.
So V=IR
V=2.14*0.25
Voltage in the 0.25 ohm is 0.535.

 

[Chi Meson]

In the last part, the question asks for the powere dissipated by the battery. This is referring to the rate at which heat is generated, so in this case you would use the voltage drop in the battery (.535 V) not the full emf of the battery. THe latter would be the answer if the question was "find the power delivered by the battery" or something like that.

It is more obvious if you use the P=I^2r version of the power equation, where r is the internal resistance.

To help with the understanding of the concepts, one should not refer to the "voltage in the resistor." THe voltage is referring to the difference in potential when comparing one side of the resistor to the other. The potential (energy per unit of charge) "drops" as the current flows through the resistor (since energy was transferred out of the circuit as heat). So we refer to the "voltage drop across a resistor."